Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<pair<bool, pair<int, int> > > mp;
int n, m;
int graph[102][102];
int ans = 0;
bool check() {
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (graph[i][j] > 0) return 0;
    }
  }
  return 1;
}
void row() {
  for (int x = 1; x <= n; x++) {
    int mn = 10000007;
    for (int y = 1; y <= m; y++) {
      mn = min(mn, graph[x][y]);
    }
    if (mn > 0) {
      ans += mn;
      pair<int, int> p(mn, x);
      mp.push_back(pair<bool, pair<int, int> >(0, p));
      for (int y = 1; y <= m; y++) {
        graph[x][y] -= mn;
      }
    }
  }
}
void col() {
  for (int x = 1; x <= m; x++) {
    int mn = 10000007;
    for (int y = 1; y <= n; y++) {
      mn = min(mn, graph[y][x]);
    }
    if (mn > 0) {
      ans += mn;
      pair<int, int> p(mn, x);
      mp.push_back(pair<bool, pair<int, int> >(1, p));
      for (int y = 1; y <= n; y++) {
        graph[y][x] -= mn;
      }
    }
  }
}
int main() {
  while (~scanf("%d%d", &n, &m)) {
    memset(graph, 0, sizeof(graph));
    mp.clear();
    for (int x = 1; x <= n; x++) {
      for (int y = 1; y <= m; y++) {
        scanf("%d", &graph[x][y]);
      }
    }
    if (n > m) {
      col();
      row();
    } else {
      row();
      col();
    }
    if (!check()) {
      puts("-1");
      return 0;
    } else {
      printf("%d\n", ans);
      vector<pair<bool, pair<int, int> > >::iterator it = mp.begin();
      int mpn = mp.size();
      for (it = mp.begin(); it != mp.end(); it++) {
        bool way = it->first;
        pair<int, int> pp = it->second;
        int num = pp.first;
        while (num--) {
          printf("%s %d\n", way == 0 ? "row" : "col", pp.second);
        }
      }
    }
  }
}